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Spreading across suits

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A game uses a deck of $40$ cards in $4$ suits of $10$ cards each. You deal yourself cards face down from a shuffled deck. What is the fewest number of cards that guarantees at least $3$ cards in each of at least $2$ different suits?

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Hint

Build the largest hand that fails: fill one suit and keep the others small.

Solution

A hand fails when at most one suit has $3$ or more cards.
The largest failing hand takes all $10$ cards of one suit and $2$ cards of each of the other $3$ suits.
That is $10 + 3 \times 2 = 16$ cards.
So the answer is $16 + 1 = 17$ .

Answer: $17$

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